K Best

Time Limit: 8000MS	Memory Limit: 65536K
Total Submissions: 11555	Accepted: 2978
Case Time Limit: 2000MS	Special Judge

Description

Demy has n jewels. Each of her jewels has some value v_i and weight w_i.

Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k

 best jewels for herself. She decided to keep such jewels that their specific value is as large as possible. That is, denote the specific value of some set of jewels S = {i₁, i₂, …, i_k} as

.

Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.

Input

The first line of the input file contains n — the number of jewels Demy got, and k — the number of jewels she would like to keep (1 ≤ k ≤ n ≤ 100 000).

The following n lines contain two integer numbers each — v_i and w_i (0 ≤ v_i ≤ 10⁶, 1 ≤ w_i ≤ 10⁶, both the sum of all v_i and the sum of all w_i do not exceed 10⁷

).

Output

Output k numbers — the numbers of jewels Demy must keep. If there are several solutions, output any one.

Sample Input

3 2
1 1
1 2
1 3

Sample Output

1 2

Source

題意：

就是最大化平均值。

POINT：

二分答案。

二分搜尋法模型：
    條件C(x):可以挑選使得單位重量的物品價值不小於x->求滿足條件的最大x->如何判斷C(x)
    價值和/重量和>=x
    價值和-重量和*x>=0
    和（價值-重量*x）>=0
    可以對（價值-重量*x）的值進行貪心的選取，選取最大的k個 和>=0

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>

using namespace std;
const int maxn = 100100;
typedef pair<double, int> pr;
int n,k;
int v[maxn];
int w[maxn];

int check(double x){
    pr y[maxn];
    for(int i=1;i<=n;i++){
        y[i].first=v[i]-x*w[i];
        y[i].second=i;
    }
    sort(y+1,y+1+n);
    double sum=0;
    for(int i=n;i>=n-k+1;i--){
        sum+=y[i].first;
    }
    if(sum>0) return 1;
    return 0;

}
void haha(double x){
    pr y[maxn];
    for(int i=1;i<=n;i++){
        y[i].first=v[i]-x*w[i];
        y[i].second=i;
    }
    sort(y+1,y+1+n);
    for(int i=n;i>=n-k+1;i--){
        if(i!=n) printf(" ");
        printf("%d",y[i].second);
    }
    printf("\n");
}
void solve()
{
    double l=0,r=100000;
    while(fabs(l-r)>1e-6){
        double mid=(l+r)/2;
        if(check(mid)){
            l=mid;
        }else{
            r=mid;
        }
    }
    haha(l);


}
int main()
{
    while(~scanf("%d %d",&n,&k)){
        for(int i=1;i<=n;i++){
            scanf("%d %d",&v[i],&w[i]);
        }
        solve();
    }

}